text:
Can we infer whether a given approach is visual only from the track data and the meteorological conditions?
I understand that even during IFR, the ATC can authorize visual approach to the pilots. Are there any tell-tale signs from track data (for example as on FlightRadar24) that will show the aircraft is being flown on a visual approach? I am trying to distinguish approaches and operations that are visual and instrument from the flight tracks in the terminal airspace.
Edit: I want to clarify that I meant instances of IFR during VMC. I believe it happens around big airports. During IMC I assume the pilots would not be able to do any visual approach.
Thank you

@757toga The OP's first sentence makes good sense if by IFR we understand an instrument flight rules clearance, which can include operations happening in VMC, while it would make little sense to discuss a clearance for a visual approach while in IMC. While there are certainly instances where writers use "IFR" when they really mean "IMC", I don't see how this is likely to be one of them.

@757toga The australian Visual Flight Rules Guide contains "Visual approach (AIP ENR 1.1) ATC authorisation
For a VFR flight by day or night ATC may give you a visual approach when you are
within 30 NM of the aerodrome. ..."

According to faa-regulations,1 for an instrument approach there is the concept of an approach gate, an imaginary point in space which is along the final approach course and is the farther of:

1NM outside of the final approach fix, or
5NM from the runway threshold.

Aircraft being vectored to intercept the course of an instrument approach procedure must be vectored to intercept final in accordance with the following:

If the weather does not permit vectoring for a visual approach, and for all RNAV approaches, no closer than 2NM outside of the approach gate;
If the weather does permit vectoring for a visual approach, for non-RNAV approaches, no closer than the approach gate;
Only upon pilot request, for non-RNAV approaches, inside the approach gate but in no case closer than the final approach fix itself.

So a tell-tale sign that an approach was a visual one is if the track turns base-to-final inside of the FAF.
Of course on many visual approaches the pilot will fly out beyond the FAF, whether due to ATC instructions (e.g. for traffic sequence) or due to company policy. Other signs that an approach was visual may include:

A continuous downwind-to-final turn instead of a defined base
Lack of a "dogleg" turn between base and final
Any track which has the aircraft flying direct to the FAF before turning toward the runway
Evidence of visual maneuvers on final, i.e. S-turns for spacing
Lack of perfect alignment with the instrument approach course on final

But it is possible for a pilot to be cleared for a visual approach and then elect to use an instrument procedure as back-up guidance, and if they do so it might appear as if they were cleared for the instrument approach.

1Not regulations as such but FAA Order JO 7110.65 which provides rules and procedures that ATC must follow.

Nice explanation! What language can a pilot use to unambiguously request a vector to intercept final inside the approach gate (i.e. as close to the FAF as possible)?

If you're volunteering it, "we'll take a vector inside the approach gate" is just fine. To be honest though the more common situation is that I misjudge when to turn someone's base, they'll be too tight, and I have to say "this vector will take you inside the approach gate, is that okay or do you want me to vector you around again?" To which the pilot will invariably reply that it's fine, they want to land sooner rather than later, thanks.

Cool. Don't recall hearing "inside the approach gate" terminology before; will try to remember that one. Guessing that "short approach" or "short/close-in vectors" aren't standard enough to necessarily accomplish the same thing?

Yeah, those probably wouldn't be specific enough for most controllers to be comfortable doing it. The "approach gate" term should get you exactly what you want though.

authors: Vladimir F Героям слава


text:
Why should the centre of mass be ahead of the centre of pressure?
Why is the centre of mass located ahead of the centre of pressure? And what would happen if centre of mass would be too close or too far away from the centre of pressure?

It isn't. The overall centre of pressure must coincide with centre of gravity to maintain attitude; otherwise the plane would rotate.

Related: http://aviation.stackexchange.com/questions/31878/what-is-the-aerodynamic-centre-and-how-does-it-relate-to-the-pitching-moment

On most airplanes, the horizontal stabilizer provides a downward lift. The center of mass is between the nose gear and the main gears.

If the center of mass is too aft:

Stall (and spin) recovery will be difficult or even impossible
Increases the chance of tail strike
The plane will tip on its tail if the main gears are not moved aft

If center of mass is aft of center of pressure, then the horizontal stabilizer must provide positive (upward) lift; the tail structure must be made stronger to support the higher lift. The wings will be moved forward of the main gears to keep the aircraft balanced on the ground. Such designs are called Tandem wing. See also this question.

Thanks for answering, but I dont really know what "aft" means, Could you clear that up?

"aft" is a fancy word for "towards the back of the aircraft". Dictionary

Oh, thanks! Sooo, what would happen if the centre of mass is ahead of the centre of pressure, but really really close to it. What are the problems that are faced when this happens?

@DevanshRathi you may check out this related question.

Why should the tail assembly be stronger if the tail provides an equal amount of upwards lift vs. downwards lift? Also, this answer doesn't really address why stall/spin recovery will be more difficult (perhaps just link to a question like this?)

@sanchises I believe why stall/spin recovery is more difficult with an aft CG is outside the scope of this question.

Comparing a design where the CoG is forward of wing CoP and a design where it is the same distance aft wing CoP, the horizontal stabilizer needs to be exactly the same size, because it will produce the same lift — just downward in the first case and upward in the second. Anyway, the main thing you need to mention is stability.

Consider an arrow. Its center of pressure is behind the middle of the arrow due to the fletching (feathers) at its back (aft) end, but its center of gravity is close to the middle. If it isn't pointing directly into the path it is taking, the center of pressure, being behind the center of gravity, pushes in the direction to restore it to pointing where it is going. Were it the other way around, the center of pressure would cause it to diverge from that direction, ultimately turning around. That's why arrows are shot with the feathers in the back, not the front.
Note that the vertical surfaces on an airplane (i.e. the vertical stabilizer and rudder) are on the back of the airplane for the same reason.

If the center of mass is behind the center of lift, then the tail wings must produce an upward lift.
This lift decreases when the plane pitches up.
This decrease in force makes the plane pitch up more which is unstable equilibrium

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text:
Do any Waco aircraft have model names?
Many aircraft manufacturers give names to the aircraft models. Piper had the Cub, Super Cub, and all the Comanche/Cherokee/Navajo/etc. names used in branding. Cessna has the Skyhawk, Skylane, etc.
Did Waco aircraft use anything other than the three-letter names for branding?
Edit: I am wondering about the original WACO aircraft, not the company that produces reproductions today.

Waco Aircraft was originally called "Weaver Aircraft Company of Ohio" in 1919. The first Weaver Aircraft Company aircraft was the 1920 "Cootie" but was it also called the "Waco Cootie". It was a monoplane that crashed on it's first flight and then rebuilt as a biplane.
The aircraft designs were called "Waco" as in "Waco 9" "Waco 10" even though the company became "Advance Aircraft Company" in 1923. In 1929 the name was formally changed to "Waco Aircraft Company".
The last Waco production aircraft design was the E Series "Aristocrat" biplane. There was also one prototype Model W "Aristocraft" which was a pusher prop, monoplane.  All the other Waco types had model designations consisting of letters and numbers, or military designations.

I wonder why this answer escaped any u-votes (or at least didn't gain a net score above zero) ever since 2-18-19?

Waco does have a model called Great Lakes.
See also: Wikipedia: Great Lakes Sport Trainer

Good point! I was actually thing more about the original WACO company, which didn't produce the Great Lakes. I'll update the question.

Down voted because the "Waco Great Lakes" is not a Waco design. Great Lakes is the name of an aircraft manufacturer who produced many different types of aircraft.   The "Waco Great Lakes" is an updated reproduction of the original Great Lakes 2T-1A-2.

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text:
What is the meaning of "altimeter three-zero-one-one"?
I've been listening to a lot of liveatc.net recently (mostly at CYVR).
What does it mean when the ATC give an "altimeter" reading (usually around the 3000 area).

Altimeter three-zero-one-one.

I'm assuming it's like a QNH setting? But not in kPa?
Addon Question:
Is inches of mercury specific to the Americas? North America or other places as well? I've only flown in South Africa where we use kPA (SI). We'd get information from tower along the lines of QNH 1013.

The units are inches of mercury.

Related and this one too, especially to your question about where inches are used

QNH 1013 is in hPa (not kPa). Generally we prefer to use 10^3 steps (10^-3/milli, 10^0/unit, 10^3/kilo, 10^6/mega, 10^9/giga, etc) but here 1013 hPa uses the hecto (10^2) prefix instead of 101.3 kPa just because many countries previously used the non SI unit bar, especially the millibar. It was easier to ask people to switch from 1013 mb to 1013 hPa, rather than to 101.3 kPa. For people using Hg inches, moving to the SI unit is more difficult.

@ymb1, many places in South America also use inHg. So the limit seems to be the Americas.

That is a QNH altimeter setting in inches of mercury, with the decimal omitted.  So in the case given, that is a QNH of 30.11 inHg or about 1020 hPa.

Ok, that make sense, Thanks. I didn't know that people were still using that measurement.

But the inch is supposed to be subdivided into powers of 2 rather than decimally, isn't it? So shouldn't it be $30\frac{7}{64}''$ instead of $30.11''$?

@HenningMakholm  No, never seen that convention in aviation in the US (or elsewhere, for that matter).  Every altimeter I've used has had decimal settings.  Some altimeters have both mb and inches, i.e. two scales that move simultaneously when the adjustment knob is turned.  But I don't think I've ever seen one that worked in x/64" fractions.  Nor heard it reported that way.

@HenningMakholm - when measuring a distance with a ruler or tape marked in inches, the inches will commonly be divided into powers-of-two units because this is traditional. However, when setting "inches of mercury" on an altimeter the setting will be in whole inches plus decimal partial-inches, e.g. "30.11".

That refers to an altimeter adjustment to compensate for local atmospheric variations in pressure.  The reading given is what the atmospheric pressure would be on the field if it were at sea level in inches of Mercury (inHg).  This is most often used in adjusting a sensitive barometric altimeter in an aircraft so that it reads an altitude which is within +/- 75ft of the actual value.  These sensitive barometric altimeters are required equipment on IFR flights and much more useful on VFR flights as well.
ATC often give an altimeter readout when handing aircraft over to another ARTCC or approach controller, generally as a courtesy to the pilot and, in the case of IFR flights, to ensure the flight crew is using the correct altimeter settings.  Other cases include at controllers at towered airports, in between the hourly ATIS broadcast update, when there has been a noticeable and fast change in the local barometric pressure at the field.
Off field and bush flying operations can make use of the same information in the form of portable weather stations such as Kestrel's series of handheld weather stations which provide barometric pressure readouts.  Pilots also will just set their altimeters to the established field elevation listed on charts in lieu of an altimeter readout as well.  The altimeter should then read to within +/-75 ft of the correct altitude, provided the unit was serviced and inspected within 24 calendar months of the flight.

It is the barometric pressure at the airfield, in inches of mercury. It is to two decimals, so you got 30.11 inches of mercury. It is the same as QNH (which is the airfield pressure adjusted to sea level). From flightsim.com

... (it) is the barometric pressure, given in inches of mercury. If you look on a standard wall barometer, you'll see that it equates to "fair". In the UK, we adopted SI units (hectopascals/millibars) some years ago, and 29.92 inches is the same as 1013mb. Anything above this value is regarded as "high" pressure, and anything below it as "low". In the days before weather forecasts were dumbed down, you'd often see synoptic charts with pressure values on them.
In Europe, again, we don't say "altimeter", instead we use what are known as "Q codes", that go back to the early days of Radio Telephony and Morse Code. So instead of hearing "altimeter 2992", in the UK you would hear "QNH 1013". Effectively it means the same thing.

Is is not the pressure at the airfield unless the airfield is at sea level.  As I write this, Telluride Regional (KTEX) in CO with a field elevation of 9069 ft MSL has an altimeter reading of 30.16 inHg.  The real pressure onsite is closer to 21.09 inHg.  The same thing applies to IFR aircraft cruising in the flight levels and using a seal level STP altimeter setting of 29.92 inHg, when the outside air pressures are less than half that value.

At least on amateur radio, with a few exceptions, using Q-codes on phone is discouraged. Basically, it's okay to state that your QTH is Springfield, but it's generally not recommended to ask for the QTR (because it's usually much easier to ask for the current time) or state that QSQ Bart Simpson (because it's usually easier to state that Bart is onboard or at the station).

Altimeter settings are usually QNH (pressure at sea level), but they're occasionally QFE (pressure at field elevation).
The international standard for pressure is hectopascals (hPa), which is indicated by a Q in METARs.  The US and some other countries use inches of mercury (inHg), which is indicated by an A in METARs.  They're also in separate numerical ranges, e.g. Q1013 = A2992, so you can easily deduce which is being used even without that hint.
Many aircraft (particularly ones expected to make transoceanic flights) will have altimeters that can be set using either unit; for those that don't, pilots will need to carry a conversion table when flying in areas that use units not matching their equipment.

authors: Bob Jarvis - Слава Україні

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text:
How can I keep my IR currency outside the USA?
I am an FAA PPL/IR holder living in Spain.
My IR license was issued in May 13 of 2016, it has been over 6 months so now I need to log my 6HIT with a safety pilot.
Is it possible to complete the currency requirements in a simulator even though I have gone past the 6 month mark without flying? Is it possible to complete the currency requirements in the EU with an instructor?
Thanks for any help, my flight school is being really unhelpful as they seem to actually know far less than I do about FAA regulations, this is my last hope!

"This is my last hope" - I would have thought contacting the FAA directly and asking them would be a suitable option..

In addition to contacting your home FSO, if your stay was going to be extended, while you were current, you might have been able to convert your license to the local equivalent through IACO.

The easiest option is to find a US CFI/AI holder in the country you are in, to sign off on your approaches.  It might be allot easier if s/he is also a Spanish CFI particularly if you are flying in a Spanish registered aircraft.  You do not indicate if you hold a certificate from Spain.
It is not uncommon to find instructors with US FAA CFI ratings in other countries.
There is no prohibition of a US CFI giving instruction in another country.  There may be of a given country, or operational limitations imposed by aircraft registration, host country operating rules, and insurance limitations, as well as owner / lease limitations.
However, I routinely sign off US pilots for approaches made at Canadian airports, in fulfillment of their US instrument recency requirements.

Yes.  You can accomplish your instrument experience in a flight simulator, flight training device or advanced training device (ATD).
An authorized instructor with FAA certificates and appropriate ratings must be present for you to log and count the instrument time for recency of experience purposes.
The only reason to accomplish instrument experience is to satisfy the recency of experience requirements in §61.57.  This experience has to be logged according to §61.51.

There are two separate types of recency of experience listed in §61.57(c): maintaining instrument experience 61.57(c)(1), (2) and (3) and to regain instrument privileges after it has lapsed for more than six calendar months.  §61.57(d) lists out what is required to regain instrument privileges.
§61.57(c) Instrument Experience

(1) Use of an airplane, powered-lift, helicopter, or airship for maintaining instrument experience
(2) Use of a flight simulator or flight training device for maintaining instrument experience.
(3) Use of an aviation training device for maintaining instrument experience.
(4) Combination of completing instrument experience in an aircraft and a flight simulator, flight training device, and aviation training device.
(5) Combination of completing instrument experience in a flight simulator or flight training device, and an aviation training device.
(6) Maintaining instrument recent experience in a glider.

Each of these has its own requirements to maintain instrument experience.  If you are still current you are allowed to keep yourself current using one of the methods listed above.

§61.57(d) - Instrument Proficiency Check

A person who has failed to meet the instrument experience requirements of paragraph (c) for more than six calendar months may reestablish instrument currency only by completing an instrument proficiency check

(1) The instrument proficiency check must be—

(i) In an aircraft that is appropriate to the aircraft category;
(ii) For other than a glider, in a flight simulator or flight training device that is representative of the aircraft category; or
(iii) For a glider, in a single-engine airplane or a glider.

If a pilot lapses with their instrument currency for more than six calendar months, they must complete an IPC check.  If they are in months 7-12 (or 3-8 for ATD in #3) they can complete the previous requirement laid in in §61.57(c).
The IPC must be completed in an aircraft, flight simulator or flight training device.  An ATD is not allowable for an IPC check.

§61.51 Pilot Logbooks

(g) Logging instrument time.

(1) A person may log instrument time only for that flight time when the person operates the aircraft solely by reference to instruments under actual or simulated instrument flight conditions.
(2) An authorized instructor may log instrument time when conducting instrument flight instruction in actual instrument flight conditions.
(3) For the purposes of logging instrument time to meet the recent instrument experience requirements of §61.57(c) of this part, the following information must be recorded in the person's logbook—

(i) The location and type of each instrument approach accomplished; and
(ii) The name of the safety pilot, if required.

(4) A person can use time in a flight simulator, flight training device, or aviation training device for acquiring instrument aeronautical experience for a pilot certificate, rating, or instrument recency experience, provided an authorized instructor is present to observe that time and signs the person's logbook or training record to verify the time and the content of the training session.

61.1. Definitions

Authorized instructor means—

(i) A person who holds a ground instructor certificate issued under part 61 of this chapter and is in compliance with §61.217, when conducting ground training in accordance with the privileges and limitations of his or her ground instructor certificate;
(ii) A person who holds a flight instructor certificate issued under part 61 of this chapter and is in compliance with §61.197, when conducting ground training or flight training in accordance with the privileges and limitations of his or her flight instructor certificate; or
(iii) A person authorized by the Administrator to provide ground training or flight training under part 61, 121, 135, or 142 of this chapter when conducting ground training or flight training in accordance with that authority.

I'm not sure that this applies to the OP, 61.51 is about logging time, not about getting current (the OP is outside the 6 month currency required by 61.57). You can accomplish some of your instrument experience in an authorized ATD, but not all of it. You can use an ATD to stay current, but I don't believe you can use an ATD to regain currency after it has expired...

@RonBeyer.  How can you prove you are current unless you log it?  You can use a ATD to stay current or get current with an authorized instructor.

Is a CFI in Spain going to count as an authorized instructor?

@TomMcW  If they have a current and valid US FAA CFI license with appropriate certificates and ratings, they are an approved instructor.

Does that mean he would have to find an FAA licensed instructor in Spain to monitor the sim time?

@TomMcW.  Yes.  That is what it means.

CFR 14 §61.57   Recent flight experience: Pilot in command. 2 allows you to use an ATD for instrument currency.
The FAA has an Advisory Circular that describes the requirements for approving an ATD. AC 61-136A Aviation Training Devices. If you know the type of device, you can go to the manufacturers website and find out if they comply with the AC.
I’ve been using a device by http://www.flythissim.com at the local flight school and can log time on it. I can’t use my home sim that uses the same software because it doesn’t conform to the AC.

I understand that I can keep it current up to 6 months, but i havent been able to find a sentence in that document which states that i can do the currency after 6 months in a simulator, because in a plane that would require a safety pilot.

"(2) Use of a flight simulator or flight training device for maintaining instrument experience. Within the 6 calendar months preceding the month of the flight, that person performed and logged at least the following tasks and iterations in a flight simulator or flight training device,…" Not sure what isn’t clear about this.

@JScarry The "within the 6 months" part doesn't apply to the OP's situation as they are outside that period. They need to take a proficiency flight, the ATD is for staying current, not getting current.

@RonBeyer He is within the 1 year time period before he requires an IPC, so there is no difference between flying with a safety pilot and using an ATD. The only thing he can’t do is fly alone in IMC.


text:
Would climbing onto fighters placed on display cause damage?
The question is about any more or less modern plane similar to Su-27 in its fuselage design. It's not uncommon to install retired planes as monuments. Here's one example in Eastern Europe (image from this post on some local Reddit equivalent):

The shot shows a retired Su-27 which had its engines and electronics removed and was installed on a pedestal somewhere in a city. It's not uncommon for locals to climb both such pedestals and the planes - youngsters do this for lulz and also to make photos of themselves. Standing and sitting on wings or behind the canopy are typical examples.
I guess this is not very good for the airplane fuselage. Not only will paint get damaged but also the fuselage may be mechanically damaged - mostly metal sheets might be bent.
I know that it's possible to accidentally fall from a plane, especially when it's wet because or rain or melting snow, but let's ignore that in this question. I'm only asking about what's good or bad for the plane.
How bad for such a plane is climbing it by an unprepared person (who doesn't know where exactly it's safe to step)?

If such planes can support several G-loads during flight and/or underwing payloads, my guess is they can deal with 80kg teens doing selflies.

More relevantly, the wing loading of the SU-27 is almost 450kg/m², so that's quite a few teens in a small space. However, your feet exert about 1000kg/m² to 2000kg/m² (back of the envelope estimate), so that's not what the plane is designed for. The skin might however be designed to allow mechanics walking on the wings rather than the wing loading.

If you intended to make the aircraft airworthy again, yes, it can cause major damage. But, for use as a static art display, it would be no different than any other display really. Most likely all the moving parts have been fixed so they don't move so making it airworthy again is pretty much impossible.

The sheet metal is not very thick, on a passenger plane the skin is only 2 mm or so. The aluminium on a fighter jet would be thicker and the spare and ribs would be closer, not sure by how much. I can only give an educated guess on this: it would probably not be easy to jump through the skin of a wing or fuselage.
It's a different matter for flight control surfaces, flaps etc. They're easier to damage by jumping on them. But the atmosphere and the weather do their job as well. While on a job in Korea I noticed a line-up of old military planes, one of them the awesome F4U. It was in way poorer condition than in this photo, had not flown in years, cockpit glass all matted, propeller stuck, engine rusted - I did a quick inspection with a thought to purchase and restore it, but the job looked too formidable. Simply from being out in the weather, unprotected, for decades.

For whatever reason, I think the F4U is the most beautiful fighter ever made.


text:
Is $24.01 million for 15 used A320 reasonable?
A few hours ago the insolvent Austrian airline Niki was bought by the owner of British Airways, IAG for a whopping \$24.01 million (plus a further \$19.08 million to provide liquidity to the company). This was the highest offer available.
The acquired assets include 15 Airbus A320 and landing rights for some European airports.
So far this was basically a summary of the Reuters article.
Given that Niki kept their business up until December 14th, 2017 it can be assumed that the aircraft are well maintained and properly working.
So summing up and getting to the question, the price seems rather low for 15 airworthy A320 aircraft in my opinion.
Did IAG get those aircraft for unbelievably low cost, or is the amount in the ballpark for such a deal?
Edit: The article also states that the landing rights are the most valuable asset of the deal, thus further devaluating the aircraft.

Update: As many comments point out, the aircraft enclosed in the deal were not Niki's own, but leased. To date I have only found German-language sources that indicate this.
Update 2: This source (as hinted by @Eike Pierstorff) clearly states, that the aircraft themselves are not included in the deal. They are in possession of Lufthansa, which in turn is bound by this document (section 5 paragraph (40)) to lease or sell the aircraft to any possible acquirer of Niki (IAG in this case) at market terms.

Apparently Niki does not own any planes (they are leased, mostly from the now defunct Air Berlin, according to various German news outlets), so saying that the prices includes 15 Airbuses seems to be a misunderstanding.

According to Reuters: "Assets include 15 A320 aircraft..." So I assumed that the aircraft were maybe somehow dissolved away from the Air Berlin assets.

Yes, I meant a misunderstanding on the side of Reuters. Tagesschau.de (biggest German news programme) specifically mentions that these are leased planes, airliners.de says the planes have actually been bought by Lufthansa as part of their failed bid. However I am just googling things, so let's hope somebody who knows the industry answers the question.

Those leases are examples of the "liabilities" (debts) that I described in my answer below. Its likely that British Airways agrees to assume those leases (i.e. continue to pay them) in exchange for the discounted sale price for the company.

Often, airlines don't buy the airplanes. They lease them. Sometimes the leases are for the commercial life of the aircraft, which means that at the end of the lease the airline basically can opt to by the airplane for a nominal amount, such as $1. Basically, it is a purchase, but it can be structured as a lease for tax purposes.

Short Answer: If the 24 Million was for just 15 airworthy A320s, then yes, its a steal because $24 Million is barely enough to buy 2 or 3 A320 engines. However, Niki Airline likely has hundreds of Millions of dollars in debt that British Airways is also "buying", that is why the final price seems so low.
Longer Answer:
There is so much that goes into the value of an aircraft that this cannot be definitively answered. Even a properly maintained aircraft may be of little value if it has too many cycles (landings and take-offs) and total hours. For example, if those 15 A320s all need engine overhauls then its going to cost a lot more than $24 Million to get those done, not to mention the downtime.
Additionally, and adding even more uncertainty to your question (sorry), is that the $24M was for the airline. As you probably know, the airline includes other assets, such as the landing rights that you mentioned, but it also includes other liabilities, such as debts to previous creditors (i.e. vendors, debt service, etc).
So, just as a simple illustration, lets say that the 15 A320 are collectively worth 150M and the landing rights are worth 200M for a total combined value of $ 350M, if the company has liabilities (debts) of 300M, then the net value would be 50M for the whole company, and paying 24M would be a steal!  But on the other hand if the company had liabilities of 400M then the company would be worth -50M. Usually, when a company is insolvent it is in this situation where it may have some valuable assets, but it has more liabilities than the assets or cash flow it generates. British Airways is likely hoping to streamline the business and turn it around, or just scrap it and take the landing rights for its own fleet.

I am aware that my question lends itself to some uncertainty, unfortunately. Factoring in the debt makes sense, and there are for sure other variables that are not known yet. Still I like the information included in your answer, if it is only to get an understanding of the costs involved in such aircraft.

@pat3d3r BTW you were right to suspect it was too good to be true, according to wikipedia the unit cost of one A320 is $99.0M (USD). So even if it is used and is worth 25% of initial cost, a single A320 would still be worth approximately 25M.

@pat3d3r It is most likely that they bought the leases for the aircraft, not the  aircraft themselves. That still means that they have to make lease payments on them for whatever terms are left on the aircraft. The $24m was most likely to pay for the equity in the aircraft leases among other tangible/intangible assets of the airline.

@Devil07 Factory list prices are way inflated, nobody pays that much. Further, quite a number of A320s have already been scrapped, the first 500-1000 made are reaching their end of life. So it's pure speculation without any knowledge of the condition of the airframes.


text:
Why does each part of span not produce equivalent downwash?
I know that it is associated with Prandtl lifting-line theory and that ideal elliptical wing actually meets the condition of my question. Unfortunately I am not good at fluid dynamics and I would like a relatively simple explanation why downwash (and lift) differs along wingspan (of non-elliptical wings). I would like to understand at least the basics and satisfy my curiosity :)

Why should non-elliptical wings not produce constant downwash over span? They do at one angle of attack, given a proper twist distribution so that the circulation distribution over span is elliptical.
However, once the angle of attack is different from that of the elliptical distribution, the change in local circulation varies in proportion to local chord. If the chord distribution is not elliptical, the new circulation distribution due to that angle of attack change is also not elliptical, and so is the change in local downwash angle.
Sometimes it is desirable to not produce an elliptical circulation distribution rsp. have an elliptical planform. If the objective is to minimise drag, it helps to reduce weight and pure lift/drag optimization is too narrow. What should count is the best ratio of lift minus wing weight relative to drag. R. T. Jones wrote a NACA Technical Note back in 1950 in which he looked at this problem analytically. Wing weight goes up when much lift is created near the tips, because this lift will cause a disproportional root bending moment, and the wing spar, which has to carry this bending moment, is a significant part of the wing structure. Therefore, reducing lift at the tips and adding more lift at the root will create a lighter wing for a modest drag increase, resulting in an overall optimum with an almost triangular lift distribution. When compared to an elliptical wing planform, the total wing span of such an optimized wing is bigger for the same overall drag, but this wing will weigh less.
Another example is wing stall. An elliptical wing without washout will have a constant lift coefficient over span, and Reynolds number effects ensure that the wing will stall at the tips first. An unsymmetrical stall will result in an accelerated roll departure: The stalled side of the wing loses lift, drops and sees an increase in angle of attack which worsens the stall condition. This effect gives an elliptical wing poor handling characteristics and is the reason why many GA aircraft have rectangular or trapezoidal planforms of little taper.

Thank you, Peter! Your answer really helps to understand the issues of elliptical circulation. However I do not understand the principle why elliptical circulation produces the least induced drag.

@RokasZilinskas: Did you read this answer already? Generally, since induced drag depends on the square of the lift coefficient, there exists an optimum and not all distributions are equally efficient.

Well, I understand that there should be the most optimum circulation because of the quadratic dependency and how induced drag is mathematically derived. But, I do not understand why each part of span of, for example, a rectangular untwisted wing, does not produce the same lift?

@RokasZilinskas: That is because span is finite. If the wing would not end, lift would indeed be equal at every span station. The lift at the wingtip has to be zero, however, because pressure will equalise around the wingtip. From there inwards lift will gradually increase until it reaches the maximum at the center. Even the center of high aspect ratio wings will never quite reach the lift coefficient of the infinite wing due to the influence of the tip.

Is it the same as saying that air below the wing spills to the top of the wing and therefore pressure equalises?

@RokasZilinskas Yes, but that is only one aspect of it. The pressure difference between top bottom gradually drops the further out you move on the wing, so all of it is affected. Maybe it helps to read this.


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Why do jet engines have a maximum speed?
What's the scientific background behind a jet engine's top speed? Jet engines increase efficiency at higher airflow rates, so why does the efficiency stop increasing at some point?

Do you have a specific example?

Remember that every technical design is a compromise between different parameters. Any specific design will have a possible speed range depending on a number of design parameters. For extremely high speed, say Mach number above 4, the engine will look very different from a typical jet engine and will most probably not generate any thrust a speeds below 1 Mach.

"Jet engines increase efficiency at higher airflow rates, so why does the efficiency stop increasing at some point?" It doesn't - your premise is completely flawed as the answers demonstrate.

Do you mean a maximum airspeed (forward velocity), or a maximum RPM of the blades?

How do you compare that to how fast you yourself can run, please?

Recommended reading here.  It seems converting turbine energy into better "mass movers" creates much more specific impulse than simply ejecting it out the rear of the jet.  (Rockets must do this as there is no air to move in space).

Because friction (aka parasitic drag).

Why might jet engines not have a maximum speed? How are jet engines different from other physical or mechanical objects with limitations?

Temperature and stability limits.
Thrust is created by accelerating a working mass in opposite direction. Net thrust is the difference between the impulse of the air flowing towards the engine and the combined impulse of burnt fuel and the air exiting the engine, derived according to time.
Impulse is mass times speed, so the exit speed of an engine needs to be larger than the entry speed to create positive thrust (assuming that the mass of the fuel in the exit stream is insignificant). Entry speed is flight speed, so the exit speed must be higher than flight speed.
The air in a jet is accelerated by heating it, so the gas expands and picks up speed in an isobaric process. The heat is provided by adding fuel, and since the fuel-air mixture in a jet is very lean, burning more fuel will produce higher exit speeds. The jet will still produce positive thrust at faster speeds.
However, adding more heat will also increase the thermal load on the turbine. By using better materials, the maximum heat in jet engines could be increased by several hundred degrees C in the past decades, but is still limited so the mixture ratio in the combustion process also must be limited. Add to that the better efficiency of higher compression, and you will have a high air temperature at the start of the combustion process already from compression, so less combustion heat can be added. That (and higher intake precompression) is the reason why modern civilian engines for subsonic flight use compression ratios of up to 50:1 while supersonic engines rarely exceed 30:1.
To get around that limitation, afterburners are used, but even here you cannot add unlimited amounts of fuel. With too much fuel the combustion process will become unstable (afterburner rumble or screech). The J-58 engine of the Mach 3.2 SR-71 even piped compressed air around the combustion chamber directly into the afterburner, so cooler air could be used to burn more fuel there. Follow the last link for more information on how the initial J-58 had to be modified to make Mach 3+ flight possible.
Ramjets run into the same temperature limit from intake compression alone: Above Mach 4 to 5 the temperature of the compressed air is so high that adding fuel will not heat the air much but will lead to dissociation as temperature increases further, using up the chemical energy to split molecular bonds instead of adding heat. Therefore, flight at Mach 5 and above requires scramjets where the air in the combustion area is still supersonic.

I believe that the questioner wants to know why we can't keep on increasing the TAS to obtain ever increasing efficiency at a constant engine fuel flow (because the efficiency does increase initially as speed increases). However, your first two sentences do answer that so +1 from me.

Maybe you could add also the contribution due to the different pressure between inlet and outlet

Or alternatively you use smart air dam design to reduce the airspeed as it enters the turbine. That's how Concorde could supercruise without afterburners at the same speed as the then-fastest jet interceptor with afterburners.

@Graham: what's a smart air dam?

@sophit How they designed the air intakes. https://www.heritageconcorde.com/air-in-take-system

@Graham: ah, the intake ramp! But that has nothing to do with the fact that the Concorde could supercruise. It had to use the afterburners at takeoff because it didn't have any high-lift system and it had simply to go fast to be able to takeoff. And it was aerodynamically optimised to produce as less drag as possible at cruise speed, so no afterburners there.

Ate you sure Concord was faster than an F104? @Graham

If a jet engine is operating at 1000 KTAS and its exhaust gases are also exiting the engine at 1000 KTAS, it is producing no thrust for the fuel it is burning - its propulsive efficiency is 0%. (If it operates at a speed higher than 1000 KTAS, it will produce drag instead of thrust - this engine is not capable of accelerating the exhaust gases to speeds any higher than 1000 KTAS, and so this speed is its "speed limit"). Turbofans achieve this limit at lower speeds, turboprops at even lower speeds.
The following equation explains why the propulsive efficiency initially increases with TAS:
$Power = work/time$
$Power = force × distance/time$
$Power = force × speed$
The force is thrust, and the speed is TAS.
Clearly, increasing the speed will increase power (while the fuel consumption remains the same). However, as mentioned earlier, thrust reduces as speed increases. This means that there will be one particular combination of speed and thrust at which their product (power) will be maximum. The speed at which this occurs is the speed for maximum propulsive efficiency, and going any faster (or slower) will reduce efficiency.
When it is said that "jet engines obtain their peak propulsive efficiency at higher speeds", it is being compared to other engines; increasing the speed doesn't necessarily mean increasing the efficiency - the efficiency initially increases, reaches it's peak, and then decreases. This is true for all air breathing engines, just that different engines attain their peak efficiency at different speeds.

To optimise fuel economy.
The jet engine needs to provide thrust that is higher than drag, the exit speed of the gas stream $V_e$ must be higher than the air speed $V_0$. For the thrust $T$ in N generated at full expansion to ambient air pressure, we can write:
$$T = \dot m\cdot \Delta V$$
with:

$\dot m$ = mass flow through the engine [kg/s]
$\Delta V = V_e - V_0$ = exhaust speed minus air speed [m/s].

The internal kinetic power of the gas stream $P_g$, provided by burning fuel, is:
$$P_g = \frac{1}{2} \cdot \dot m \cdot V_e^2$$
To generate a given amount of thrust, if we choose $\dot m$ as high as possible and $V_e$ as low as possible, we have generated the required thrust with the minimal amount of fuel required, since the $V_e$ term occurs squared in the required power equation. So we want to choose the jet exhaust speed as low as possible above the required cruise speed.
The jet engine itself does not care too much about how fast the aeroplane is flying, as long as the intake air is slowed down sufficiently and the exhaust air accelerated to above airspeed. For this we need to consider the jet engine as the core of the total engine installation.

The main design parameter for the type of jet engine to be used is the cruise speed. The image above (ref) is from prof. Wittenberg's college handout VTH-D 14, and depicts seven types of jet engine:

Ramjet
Pulse jet
Two types of turbojet
Three types of turbofans

If the aeroplane is to fly hypersonic the ramjet is the most suitable type, no turbine required since the incoming air is compressed adequately by the intake. However, the ramjet does not work from a standstill.
Turbojets and bypass turbofans are used in supersonic aeroplanes: most thrust is generated by the hot air exhaust, which exits at supersonic speed.
Turbofans are best for high subsonic aeroplanes: most of the thrust is generated by the fan, some by the hot exhaust.

The role of the intake and exhaust in thrust generation as a function of airspeed can be seen in picture above, earlier used in this answer. What is not shown here is the fuel economy: bypass & fan thrust is way more economical than exhaust gas thrust, but can only be used at lower speeds. The lower the cruise speed, the higher the fan thrust proportion should be. And fan thrust is subject to a speed limit, and above certain cruise speeds the intake-exhaust provides all the thrust.

Several points limit the top speed of the jet engine

Efficiency equation you're probably referring to is: $\eta_p = \frac{2}{1 + \frac{v_9}{v_0}}$

This never does stop increasing
Yet the growth is logarithmic with speed ~0.4*ln(v0)
Trying to fly 2x your speed only gains 1.33x efficiency
This type of efficiency only measures how much of the engine's fuel burn effort is applied to the airframe, rather than just moving air.

This efficiency matters, but it also needs Specific Impulse

Like others have noted, its not just an issue of fuel burn work efficiency
The engine is translating a certain amount of the fuel burn into work on the airframe.
How much of that work then gets applied as acceleration?
From Wikipedia: Specific impulse (usually abbreviated Isp) is a measure of how efficiently a reaction mass engine (a rocket using propellant or a jet engine using fuel) creates thrust.
Specific Impulse pretty much uniformly goes down with speed.

By Kashkhan at English Wikipedia, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=10086744

Together, you end up with curves like the below for turbofans

Jet engine design drivers: past, present and future

Finally, all these considerations also ignore drag

The airplane is flying through air
The air has density and inertia.
To put the airplane where the air used to be you have to do something with the air
Generally means diverting or deflecting the air, or piping it through the plane

Busemann Biplanes are interesting variation on the last idea, although they don't produce lift.
Drag is generally thought of as two terms (Lift Drag and Parasitic Drag)
Lift Drag, drag because you're redirecting airflow to create lift, conveniently goes down with speed.
However, Parasitic Drag, drag because the airframe shape is diverting / deflecting air and the air has a friction effect on the skin, effectively rises as $V^2$
You put them together, and you get a drag curve that looks like this:

By Olivier Cleynen - Own work, CC0, https://commons.wikimedia.org/w/index.php?curid=51974460

This is one of the main reasons that most airplanes have a cruising speed.

Ideally, what you would want (if it was not highly complex) is an engine that acts like a turboprop, then a turbofan, then a ramjet, and then a scramjet.

Move a lot of air slowly at low speeds, and then move less air with a much higher exit velocity at high speeds.

Here is a thrust equation from the NASA website, which seems to have perked up a bit since the success of the Artemis mission.
We see thrust is produced by net momentum change and net pressure differential from a given exhaust area.
With turbojets, higher pressure is created by compressing ambient air.  Heating it (by the relationship of Pressure and Volume being proportional to Temperature) allows both the exhaust area to be larger than the inlet and the exhaust velocity to be greater, producing net thrust.  There is only a slight increase in mass flow in the engine core (but adding fans or props greatly improves this in subsonic flight regimes).
The issue is that drag is created by the inlet and temperature increase is caused by compression, robbing the engine of its net thrust until a thrust/drag breakeven point is reached.  This is why some people claiming "hypersonic" aircraft these days are simply flying horizontal rockets, which are unburdened of the task of compressing adequate amounts oxidizer/mass from the atmosphere.  But, then again, that is exactly how the X-15 managed hypersonic speeds back in the 1960s.
The limiting factor for jet propulsion is inlet temperature.  Use of cryogenic fuel and/or oxidizer to cool inlet air is one way of improving overall efficiency.


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What specific hazard is the practice of identifying an ILS LOC by Morse code trying to prevent?
This question got me wondering about the practice of tuning a navigation frequency, selecting it on the audio panel, and then identifying it by listening to the three letter Morse code signal.
I know it is a sound practice, and the reason may have been explained to me years ago in a way that made sense at the time, but my understanding has been lost in the sands of time and I have gotten lazy at doing it regularly.
To clarify, I am only interested in answers that can outline a plausible scenario where a mishap could ensue as a result of not identifying an ILS/VOR station by Morse code.  For example, how an aircraft might lock onto something on the same frequency that is convincingly similar enough in the cockpit indications, that undetected deviations by the pilot could drive an aircraft unto an unsafe condition and possibly result in a mishap.
So, what is the worst thing that could happen if a pilot failed to ID a Navaid?  (Safety of flight, busting a checkride doesn't count!)
P.S. Bonus points if you reference a CFR part that requires it.
ADDENDUM:
Thanks to those who took the time to answer.  There are some good and correct answers, but a few missed the point in my asking about actual, specific risk scenarios.  Because I would never argue against the practice, and I know why it's done, it just feels a bit like an inefficient anachronism when there are so many other better ways to catch an error.
I don't have near the flight hours of a seasoned part 135 or 121 pro, but I have never once uttered in the de-brief:  "Gee, good thing we listened to that Morse code ID or we might have ____"  (Insert scenario)
And I have debriefed plenty of mistakes, my own included.  But any time I've switched navaids on an airway there is an expectation of what the next course and DME will indicate.  When that needle swings reassuringly to the inbound course, and DME is plus/minus a mile, flicking that audio switch to hear a few quick beeps feels like a casual afterthought performed out of rote habit.
Because if it locks on something 90 degrees off, and many miles diferent than expected, the first thing I do is doublecheck the entry.  Because my eyes can detect a mistake far quicker and more reliably that what it takes to process the audible code.  (I'm not steeped in telegraphy!)
I just can't think of a plausible situation where I would be fooled or otherwise miss a mistake and be saved by the Morse code.
However, a signal from a station undergoing calibration that might be picked up but is not accurate, is a very valid concern.  And that was the answer I accepted.

Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed.

Identification serves three purposes,

to check that the transmitter is actually on
you have tuned the correct frequency, and are picking correct signal
the navaid is serviceable

modern avionics are actually displaying the ident of the naviad on screens. So if you tune in a VOR or ILS for example, the system actually listens for the morse code and displays it on PFD or navigation display. Checking the ident displayed on the screen serves the same purpose as listening to the morse code. It is not too uncommon to see "TST" or "ZZZ" or some other bogus ident displayed during cruise as some FMS auto-tuned VOR underneath is being under maintenance.

The audio feature of VOR and ILS is a remnant of the 1950s. Back then, radio receivers had only an analog tuning knob for frequency selection, as can be seen on  the Narco VHT-2 Superhomer depicted below. You would coarsly tune it to the intended frequency using the scale on the top right and then listen to the audio channel and fine-tune until the signal was strongest. Since the frequency scale was not very accurate, having a morse code identifier helped to ensure you had the right frequency.
The specific hazard is "someone with a 1950s radio might tune the wrong frequency." There may or may not be reasons for listening to the audio channel with a modern radio, but compatibility with old (then contemporary) radio receivers is why these features exist in the first place.

A VOR's signal is, conceptually speaking, pretty simple. It has no authentication or error-checking features beyond the Morse Code identifier. Thus, it's entirely possible for another transmitter to send a signal that looks enough like a VOR to fool the receiver, which will then show some random course based on the garbage signal it's receiving. Now, granted, I can't think of a single type of transmitter that would be close enough to a VOR's signal for this to be a real risk, but that doesn't mean that such a transmitter doesn't exist.
Well, actually, that's not quite true. I can think of one such transmitter: a different VOR. VORs on the same frequency are supposed to be far enough apart that they don't interfere with one another. However, radio range is dependent on atmospheric conditions, and with the right conditions, radio waves can propagate much further than they were intended. (And that's assuming you tuned the correct frequency for the intended VOR in the first place. If you mistype the frequency, you may end up tracking a VOR other than the one you wanted.)
As for what can happen? The pilots of American Airlines flight 965 accidentally selected the wrong navaid on their way in to Cali, Columbia, and wound up flying into the side of a mountain.

AA965 used RNAV (DIRECT-TO) function of FMS to navigate themselves to  the mountain. They chose the nearest "R" (which is the usual case), which happened to be the incorrect one in their case and let the FMS navigate through the mountain.

Your last sentence in the first paragraph capture the sentiment that inspired this question!  On the topic of locking onto another navaid of the same frequency though, please see in my addendum why I think this is very low risk.

Strictly speaking, the answer to your question is that listening to the Morse code ID of the ILS or VOR won't save you from any specific common hazard.  But, I should point out that you are neglecting an important safety principle: redundancy.
For countries that drive on the right side of the road, why should drivers put their left turn signals on in the left-turn lane?  Because doing so assures other drivers that one actually intends to turn left.  The redundant information that 1) the driver is in the left-turn lane, and 2) the driver's left-turn signal is on, confirms the driver's intent to turn left, which reassures other drivers.
Suppose that you were entering a frequency into your ILS or VOR, but you accidentally transposed two digits.  Or maybe there is a blob on the chart that makes distinguishing a 6 from an 8 difficult.  Or the frequency reminds you of the last four digits of an old telephone number you had, but isn't the same, and you punched in your old phone number instead.  Etc, etc, you get the idea.  If for some reason you entered the wrong frequency, then if you were to listen for the Morse code ID, or read it off the display, then you would realize that something is wrong and investigate; the redundant information would help you.  If you were to ignore the Morse code ID, then you'd be ignoring redundant information that could help protect you from the possibly disastrous consequences of a simple mistake.


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Is there an FAA reference for rounding off of traffic pattern altitudes?
The convention by default is to fly the downwind leg at 1000' above ground level, but most pilots cannot really discern or fly within just a few feet of a chosen altitude, and generally prefer to fly at altitudes in hundred foot increments.
So, for example at KPAE where the field elevation is published as 606' I would target 1600' MSL as a traffic pattern altitude, vs trying to hold exactly 1606'.
Is there is any written guidance from the FAA, regulatory or not, (i.e. AIM) that describes allowable rounding off?  Or, by a specific example such as I've shown above, it's implied that TPA is always the elevation to the exact foot, plus 1000'?

Might be a systems thing where autopilots and altitude bugs only have steps of 100 feet. I know flight plans are also similar with the 100s digit being the lowest unit you can enter.

@SMSvonderTann, just to clarify, I'm not asking about bug settings.

I doubt there is a reference but the only place it would be is the TERPS (FAA order 8260)


text:
What use was there for Boeing to include customer numbers in their aircraft model names, and why did they end this practice?
Inspired by this question (thanks Hot Network Questions!).
The tl;dr is that Boeing included a unique numeric model number designation, referring to the customer that had purchased a particular plane, in the actual model number of the plane. For example, a 747-8 for Lufthansa would be reflected as "747-830", where 30 is Lufthansa's unique customer number.
What is the utility of this "customer model number"? I assume it was used downstream by e.g. fitters and painters to determine what specific customisations should be applied to the airframe, but it seems that since 2016 this practice has been phased out.
Does anyone happen to know the history behind these numbers, i.e. how and why they came into being and why Boeing chose to stop using them after literally decades?

In some other answer around here it was mentioned that Boeing used to do quite a lot of customizations even to the point that the cockpit layouts differ between some customer variants. It was later realized it was not so good idea.

@JanHudec I think this was in a comment by Terry, referring to the direction the overhead switches worked.

It's an easy way to know what configuration options "this" aircraft has, long after it has left the factory.
Let's say that airline 1 ordered a bunch of 737's with one cockpit jumpseat & two aircraft batteries and the 3-hour cargo fire suppression option -- stuff you might expect for overwater flights.  They got, hypothetically, the 737-7E1 (just making that up.... a 737-700 with customer code E1).  Then airline 2 ordered a bunch of 737-700's with two cockpit jumpseats & a single aircraft battery and the 1-hour fire suppression system... a more short-haul sort of fleet.  And let's say that their code was the 737-7E2.  Years later, all of those aircraft from both airline 1 and airline 2, and perhaps others, have been sold to airline 3.  Airline 3 has some 737-7E1's and some 737-7E2's.  And if you're conversant with the codes in that fleet (as mechanics & others at Airline 3 would be), then you can see on a list that "this" aircraft is a -7E2 and so you know that it's the configuration with both jumpseats, one battery, and etc.
Those customer codes don't deal with things like the passenger cabin seating configuration; those can be swapped out pretty easily.  But stuff like batteries & cargo fire bottles and plumbing for an extra O2 mask in the cockpit doesn't tend to change, so those sorts of things get captured in the customer codes.
As to why they abandoned them, my only guess is that they may have decided to standardize all their product by removing customizations -- a 787 is a 787 is a 787, and (perhaps) they don't let customers configure them in ways that, down the road, may make the aircraft less desirable for somebody looking to buy or lease a fleet that doesn't have lots of variants between jets.
There are stories from guys who flew freighters where ever aircraft on the ramp was different in some way -- even though they were all the same type.  Newer INS, older INS, different flight directors, sometimes even different types of engines, etc etc.  For a freight operation looking for any aircraft they can get cheap, that lack of standardization may be no big deal. But for other operators, not having to deal with the headaches of dissimilar aircraft in the fleet may be an incentive to choose "this" type over "that" one.

For the 737 MAX airlines can decide if MCAS is hooked up to one or two AoA sensors (for an upcharge). Customer-specific layouts are still a reality at Boeing, and a way to make more money.


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What requirements aren't met, that would designate an airspace to be Class E instead of A, B, C, or D?
Quoted from Transport Canada "Class E airspace is designated where an operational need exists for controlled airspace but does not meet the requirements for Class A, B, C, or D."
Class A airspace requires you to be at a certain altitude to actually be in Class A airspace.
Class B is the same, Class C requires ATC clearance for both IFR and VFR.
Class D just requires VFR to have dialogue (two way communcation) with ATC.
What requirements wouldn't be met in a particular airspace that would need to designate it as Class E instead of the other classes?
EDIT: After some further research, the easiest way I've thought about Class E is an area that requires some control strictly for IFR traffic, but to VFR aircraft it's pretty much uncontrolled airspace.
Thank you!

The simplest answer is that Class E doesn't have a control tower, and the others (Except A) do.
Think of Class E like it's "Class A", but just at the lower altitudes. Both A and E are basically "wide swaths of airspace, not really near a major airport" with control provided by the Area Control Center (similar to ARTCC in the USA). Whereas B, C, and D are designated at airports with Control Towers. The only real difference between A and E is that no VFR traffic is allowed to enter Class A.
Although you're right that VFR traffic does not need to communicate with ATC in Class E, be careful about using the phrase "to them, it's uncontrolled airspace".  It IS still controlled airspace. VFR traffic does not need (nor sometimes want) to be controlled, but that does not make the airspace "uncontrolled".  "Controlled Airspace" is a term one will likely encounter on exams, and one doesn't want to get into the habit of calling it "uncontrolled" just because one is a VFR pilot.


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CAP 413: Callsign-first replies in UK RT procedure?
Taking a UK Flight Radio Telephony Operator Licence course and have fallen into a rabbithole.
From CAP 413 edition 23 (the CAA's Radiotelephony Manual):

2.48
The placement of the callsigns of both the aircraft and the ground station within an established RTF exchange should be as follows:
2.49
Ground to Air:

Aircraft callsign – message or reply.

Air to Ground:

Initiation of new information/request etc. – Aircraft callsign then message
Reply – Repeat of pertinent information/readback/acknowledgement then aircraft callsign.

So an exchange started by the ATCO begins and ends with the aircraft callsign:

G-ABCD, runway two seven, cleared for take-off.
Runway two seven, cleared for take-off, G-ABCD

However there are counterexamples in the manual:
5.4 (page 161 in the PDF):

G-CD, report heading.
G-CD, heading three five zero.
G-CD, for identification turn left heading three two zero degrees.

5.9, one of the examples:

BIGJET 347, confirm transponder operating.
BIGJET 347, negative, transponder unserviceable.

5.25:

G-CD, traffic 11 o'clock, ...
G-CD, traffic in sight

Followed by numerous examples of "[Callsign], Roger". There's more of those in reply to an AFIS in figure 43 (PDF page 352).
Are these just inaccuracies in the manual? I spotted a couple of definite typos so it's not impossible. Or are they holdovers from a previous standard? Or, are there situations where the pilot should begin their reply with their callsign?
(Beyond the QDM procedures where the pilot starts and ends their transmission with their callsign, presumably to give more time for a fix to be taken.)

Welcome to aviation.SE! This seems like a dupe of this question, although perhaps yours is more specific.

Cheers @Pondlife, I did see that question, but that one is not specific to any country and the accepted answer quotes section 2.49 which is where I was starting from.

That's a very good analysis! If I were you I would contact the CAA to clarify any doubt for you and the public (me included!)

The important word in 2.49 is "repeat".  The callsign comes at the end of the transmission when the pilot is repeating information from ATC - in the example from 4.28, the pilot is repeating the takeoff clearance, so ends the transmission with the callsign.
In all other examples, the pilot is providing new information (the altitude, transponder status, etc) to ATC, so begins the transmission with the callsign.
[Callsign], Wilco is not a valid transmission, and does not appear anywhere (that I can find) in the current version (June 2020) of CAP 413. [Callsign], Roger is a valid transmission - the pilot is informing ATC that the transmission has been received and understood.

2.49.2 is "repeat of pertinent information, or readback, or acknowledgement" surely? You don't repeat an acknowledgement. I think I was mistaken about "[callsign] Wilco" but why would that be invalid and "[callsign] Roger" valid?

@JackDeeth Yes, that could be an alternative parsing of the sentence - put the callsign at the end if the transmission is either (a) a repeat of information, (b) a readback, or (c) an acknowledgement.  The result is the same - the transmissions which provide new information to ATC do not come into one of those three categories, so start with the callsign.

@JackDeeth For "Roger" and "Wilco", "Wilco" is an acknowledgement, so the callsign goes at the end, "Roger" is telling ATC something they didn't know (that you have received and understood their transmission), so the callsign goes at the beginning.

However, CAP 413 generally does state "Roger, [Callsign]" too…


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Beside payload size, what precise factors are considered to choose the fuselages shape?
I am currently reading this very interesting book, which explains -I guess- most of the things you need to know in order to design aircraft.
Yet it is not clear to me besides the payload size what other factors are precisely used in order to determine the exact shape and size of the fuselage. I know that aerodynamics are considered, but that's very broad...
I read that the conic lofting method is usually used to determine the exact shape of the fuselage. But I would need some more explanation to get a better understanding of how this is determined.
In other words: I know my wing size, airfoil, payload size and would like to design the fuselage. How should I proceed?
Thanks
EDIT: Feel free to share the pragmatic approach
What bothers me is that I ended up finding a webpage with some interesting guidelines, but contains many constant values I have no I idea where they cone from: http://www.barnardmicrosystems.com/UAV/uav_design/guidelines.html
eg of such values: wingspan: 1.041, 0.382; step 3 length: 1.775

What type/size of aircraft? Related: Chapter 6 of Design Of Aircraft or two thesis at TUDelft.

@mins type: fixed wing, low speed single engine. Ideally something along those lines: http://blog.tmcnet.com/robotics/Lockheed%20Martin%20Desert%20Hawk.jpg But maybe with the prop in front. I think having it at the fron may maybe make some stuff easier to understand and find out for me

Assuming an engineering background, here’s a short list of factors affecting fuselage shape and size.

aerodynamic considerations
volume considerations
structural and elastic considerations
operational and interface considerations.

Aerodynamically, drag and stability are the major concerns.
Volume wise, payloads, subsystems, cargo, propulsion, etc must be considered. So it might take a few iterations for a preliminary design to converge at a feasible and optimum solution.
Structurally, elasticity of the fuselage is important, at least for weight efficiency, and general comfort. a bad shape can cause unwanted structural modes, which could risk failure in flight, or may cause difficulty in controls. If you were ever on a 747, you probably noticed how much it snakes towards the rear of the plane. The oval shape of the fuselage has parts to blame.
Finally, sometimes doors, openings, and structural interfaces dictate geometries of fuselages.
Additionally, geometric constraints such as tail strike prevention, or belly landing can be other reasons to shape the fuselage.

I have an EE background. What good approach (techniques, alorithms, intuition, ...) would you suggest to look at if I want to investigate on point nr 1 in my current situation? The book I am reading doesn't AFAIK explain a generic way of investigating #1

Pages 109, 149 and 155 seem to cover the generals of the topic. The shape of the fuselage cannot be handled isolated from the general configuration of the aircraft. But for a small and not so expensive process I’d say go ahead and try to fit things inside and try to make it as small as possible. However, if you think it professionally, get an aircraft designer in the team so that your aircraft will (hopefully) not deconstruct mid air, due to a gust or due to some control inputs. :)

indeed it is a very general and imho not pragmatic explanation. In my ideal world the book would contain smth like: step 1 do xyz because, step 2 do yxz because etc... Is there any book you'd recommend me reading in order to know how that's done professionally? Everything seems to be proportioned so there might be a way for me to get a proper understanding of everything without going too deep in aerodynamics, strength analysis and all that stuff...

Basically I could make my payload fit in the fuselage, but I guess that for aerodynamical reasons and balancing I may have resize it a bit in order for the UAV to fly, which is why I think just making sure that the payload fits is probably only 20% of the work. So far I only have a wingspan of up to 1m (because that is the easies to build myself) and I have to size my fuselage to that. Yet according to what I found online there is no ratio between the fuselage size and wingspan/area. Which is why I'd need some guidelines


text:
How to define the aileron's deflection angle (in the Laplace domain)?
Very simple question here. I know what Laplace transforms are, but except for solving ODE's I haven't really used them in a real project or thought about it in a very pragmatic way.
I am currently reading "Small Unmanned Aircraft : Theory and Practice" from    Randal W. Beard, Timothy W. McLain and am a bit lost on how exactly I should incorporate/use a specific equation.
My question is about this equation that relates the aileron deflection to the fixed wing UAV's roll angle:

where:

It seems like the aileron's deflection angle $\delta$ is a function of s, ie $\delta(s)$.
So my first question is: suppose you want to imply the aileron rotated 45degrees how does that translate to $\delta(s)$?
Secondly, I imagine I should let s vary from 0 to infinity, that will give me a graphical representation in the s domain, but that I think won't tell my how many degrees my roll angle is relative to some axis. Or am I missing something?
EDIT:
Another link to the book, with mirrors (libgen.io, libgen.pw, bookfi.net, bookzz.org) to its content depending on your location: http://libgen.io/book/index.php?md5=7314182B194BAB33173B521614B42663
I am trying to achieve latteral tracking and/or terminal control on a fixed wing UAV for autodidactical purposes.

I could not open the book you reference from your link and therefore cannot see the context of the equation.

@Koyovis added a link at the bottom of my post

It looks like a transfer function for a feedback loop, which is a function of time and transformed into the frequency domain. Input is the aileron deflection $\delta(s)$, output is the roll angle $\phi(s)$. When deflecting the ailerons, the aircraft is accelerated around the X-axis. The roll movement creates a damping force proportional to span $b$, which quickly dampens out the roll acceleration into a steady roll rate.
So achieving a roll angle depends on the moment of inertia around the X-axis, and wing area geometrical data, plus the time that the aileron is deflected at a certain angle. In order to get to a specific bank angle, the aileron needs to be deflected, then returned to zero before the bank angle is achieved - damping and inertia will then continue to roll the aircraft until the target bank angle is achieved.
The transfer functions defined in the frequency domain are used for determining gain and phase shift, more info for instance here, section 6.3 onwards. Using a signal in the frequency domain gets to be a pain once the deflection is a real world time history and not a mathematical function like a sine wave or a pure step input.
Image source
Once the gains are determined with the Bode plots, the block diagram of the transfer function is pretty useful for determining bank angle as a function of the time history of aileron deflection: just read the actual bank angle, subtract from the desired bank angle, and set the aileron deflection angle in real time according to the difference signal.

EDIT
Have managed to have a look into the book. I'm familiar with the equations, describing stability and control of aircraft. Your question on how to define an input signal in the frequency domain is a pertinent one: it would have to be defined mathematically in the frequency domain as well. I have only seen use of the frequency domain equations in off-line analysis, not in real-time use since the inputs are never purely mathematically defined. Equations such as 5.53 for the time domain. Then any input signal can be added in real time, multiplied with the gains found from the Laplace transform, integrated and output.

So my first question is: suppose you want to imply the aileron rotated 45degrees how does that translate to δ(s)?

It depends how fast you bring the ailerons to 45º, for how long you keep them there, and what happens next. If it is a step response, the function is $\frac{0.785}{S}$. The aircraft will respond with a steady roll rate (in aeroplane axes) and will never establish a constant bank angle.

You probably couldn't open the links due to your location. Hadn't thought about that :), I added a link to a webpage with multiple mirrors in order to have access to the book's content.

If I'm not mistaken according to the book this is an open loop control system (block diagram: 5.2 page 69). You say TFs  are used for determining gain and phase shift, that's indeed what I learned while being at uni. But how is this going to be useful here? How do you suggest me to proceed with this equation (or another one in this book) to know how much I need to deflect my ailerons to get a certain roll angle?

How much and how long for, given the dynamics of the aircraft and of the aileron deflection. The new link shows the book behind a pay wall, sorry still cannot see the block diagrams and the context. I've never used the frequency domain equations directly in real-time code, only equations in the time domain.

I can try to share the book this way: https://www.dropbox.com/s/ueiuu7ulhn2h7ti/Randal%20W.%20Beard%2C%20Timothy%20W.%20McLain-Small%20Unmanned%20Aircraft_%20Theory%20and%20Practice-Princeton%20University%20Press%20%282012%29.pdf?dl=0
I personnaly think it is a very comprehensive book.


text:
Why is the velocity vector of this rocket not aligned with the rocket's axis itself?
I am reading a book about rocket guidance. And came accross this drawing:

The missiles axis is always oriented to the moving target with this guidance system. But I was wondering, knowing the exhaust nozzles and thrusters are at the back of the missile, how is it possible to have the velocity vector (which I indicated with a red circle) to be oriented vertically like that? I was expecting this vector to be aligned with the longitudinal axis of the missile itself so that the angle alpha (angle of attack) equals 0.
Seems like, beside this one, there are other guidance methods where we try to by all means always try to keep the angle of attack equal to zero (chase guidance). What is the difference?
Entire text: http://paste.ubuntu.com/25706952
Original text: https://i.sstatic.net/3oGgL.jpg

This is almost never true for any missile. In this particular case it's called "Proportional navigation". I know too little about it (or more accurately, forgot too much about it) to form an answer, but hopefully that key word would lead you to more helpful materials.

This is a schematic drawing the rockets orientation is unrelated to the velocity vector.
The drawing shows angle of the rocket at the start of its flight, which it would follow without the guidance system (left straight line) and the (theoretical) angle the rocket could have been launched at to hit the plane at the "meeting point" (lower straight line).
The angle of the rocket in this drawing is along the direct hit line most likely to symbolise it is aiming at the plane...

I don't think they would have specified that the angle alpha (angle of attack) is needed.  – LandonZeKepitelOfGreytBritn

despite what the text may or may not say:

alpha is not the angle off attack.
The  angle off attack is between the plane and the rocket at the moment of impact.
alpha (as shown in the drawing) is the difference in angle between the real rocket launch angle and the angle between the launch facility and the plane in the moment of impact.
It may be that this is referred to as the angle of attack in military speech but that's wrong in the physical sense.

alpha is very much the angle of attack. it is the angle at which the body "attacks" the air, i.e. the angle between the forward axis of the vehicle and the direction of the incoming stream of air. The "moment of impact" has nothing to do with it. – Federico

You are right with your definition of angle of attack as we use it in aviation.
But the drawing has no indication of the "the direction of the incoming stream of air* (other than implicit from the rockets movement) and even if it would it would not relate the angle alpha as it is drawn since this would be unrelated to the targets movement. But in the drawing the angle alpha is related to the targets movement.

not sure this is correct. Because the text states this: https://goo.gl/oahd4h , if it were for aesthetical reasons or only just for symobics I don't think they would have specified that the angle alpha (angle of attack) is needed. Entire text: http://paste.ubuntu.com/25706952/ (can't create a short link for the translated version, you may have to translate it yourself), original text: https://imgur.com/a/CfWDb

"the drawing has no indication of the the direction of the incoming stream of air" well, as you say "other then implicit from the rockets movement". Usually on textbooks, unless required, you neglect the wind, so in that diagram the direction of the incoming stream of air is defined.


text:
Is this calculation for the 'free' climbing speed when flying in a headwind with a gradient correct?
From energy considerations, it can be found that a plane flying straight and level, if it enters a headwind with a positive gradient, will climb, extracting energy from the wind gradient.
The extra energy involved will be
$E_{EXTRA}  = mg(h_f – h) = \frac{1}{2} m (vfg^2– vg^2) $
where $m$ is the mass of the plane, $h$ the initial altitude, $h_f$ the final altitude after the climb, $vfg$ the speed of the plane (when starting the climb) in relation to the mass of air present at the altitude where the climb will end, and $vg$ the TAS of the plane when starting the climb.
Now, the question is what the extra power of that airplane will be when climbing under those conditions. It’s clear that there is an extra power, since energy is being taken from the wind gradient at a given rate…
As that rate is related to the magnitude of the gradient, we may say that, for a given wind gradient $ß$, the extra power $P_{EXTRA}$ will be:
$P_{EXTRA} = E_{EXTRA} · ß$
$P_{EXTRA} = \frac{1}{2} m (vfg^2 – vg^2) ß$
Knowing the power, we can now find also the extra vertical speed w of the plane of mass m due only to the gradient:
$w = \frac{1}{2} m (vfg^2 – vg^2) ß/mg   =  (vfg^2 – vg^2) ß/2g$
In order to 'test' the formula, we may think that the difference between the wind speeds at the top and the bottom of a 'block of wind' is $12 m/s$, that we fly with a TAS of $vg = 60 m/s$ and that $vfg = 72 m/s$. We set the gradient at $10 m/s/100m = 0.1 s^{-1}$. Thus, the height of the ‘wind block’ will be 120 m.
Then,
$w = (722 - 602) · 0.1/2 · 9.8  = 8.08 m/s = 1591 feet/min$
Not implausible, I think…

I propose to look at it in a different way. Let's assume a constant airspeed climb.
We have two scenarios: one where there is no headwind gradient, and one with a headwind gradient. We can calculate the achieved climb speed in both scenarios and then compare them to see what the effect of the wind is. By selecting a constant airspeed, the drag is constant in both scenarios so they can be easily compared.
In both scenarios the aircraft will have the same constant thrust, such that the thrust exceeds the drag. The excess thrust is equal to 5% of the weight.
For the purpose this answer, I do not look at calibrated airspeed but only at true airspeed. Basically I neglect effects of changing density with altitude.
Also, the angle of climb (flight path angle $\gamma$) is sufficiently small to be in the linear domain (such that $\cos\gamma \approx1$ and  $\sin\gamma \approx \gamma$)
Scenario 1: climbing in a constant wind field
In the first scenario the aircraft is climbing in a constant wind field. This can be no wind at all, or a constant headwind or a constant tailwind.
For the aircraft to perform a constant airspeed climb in this constant wind field, all forces acting on the aircraft (thrust, drag, lift, weight) need to be in equilibrium.

Thrust = green
Drag = red
Lift = blue
Weight = black

$\frac{dV}{dt} = \frac{T - D - W sin \gamma}{m} = g\left( \frac{T - D}{W}  - \sin \gamma\right) = 0 $
With a thrust excess ($T-D$) of 5% of the weight, the aircraft need to climb at a flight path angle of 0.05 radians which is approximately 2.86 degrees.
$\frac{T - D}{W} = \sin \gamma = 0.05$
The vertical speed $w$ is then equal to:
$w = V\cdot\sin\gamma $
So in the first scenario the aircraft will achieve a climb speed that is equal to 5% of the true airspeed. Taking 60 m/s from your calculation for the airspeed, the aircraft will climb at 3 m/s = 591 ft/min.
Scenario 2: climbing in a wind gradient
In our second scenario, the same aircraft climbs with the same velocity and same thrust setting, but now it encounters a headwind gradient of 10 m/s per 100 meter of climb. This equates to a wind gradient of 0.1/s:
$\tau = \frac{10 \textrm{ m/s} }{100 \textrm {m}} = 0.1 \frac{1}{\textrm{s}}$.
A climb of 3 m/s would result in an effective wind increase of $\tau \cdot w$ = 0.3 m/s2 which is effectively an increase of true airspeed. Therefore, this isn't a constant speed climb anymore.
To make sure the climb is performed at a constant air speed, the aircraft needs to climb faster.
$\frac{dV}{dt} = \frac{T - D - W sin \gamma}{m} = g\left( \frac{T - D}{W}  - \sin \gamma\right) + \tau V \sin \gamma = 0 $
$\sin \gamma =  \frac{g}{g-\tau V}\cdot\frac{T-D}{W}= \frac{9.81}{9.81-0.1\cdot 60}\cdot0.05 = 0.129$
$w=V\sin\gamma =$ 7.72  m/s = 1521  ft/min

In the wind gradient the climb speed of 1521 ft/min is considerably higher than the 591 ft/min achieved in a constant wind field with the same power setting. The difference is a factor of 2.57 which can be attributed to the wind.

Thanks for your valuable comments. However, and in 'scenario 2', I think that the climb is 'unintended'... I mean that the pilot is flying s/l, in non-moving air, then suddenly the wind conditions change to a headwind with gradient, and then the aircraft starts to climb automatically, and 'entirely free'... I mean that no fuel is spent on the climb, all the energy gained in the climb coming from the wind gradient...

@xxavier I'll think about that scenario and see if I can add it. The problem is that as soon a you start climbing, the lift vector is tilted backwards so the aircraft will decelerate. Of course kinetic energy can always be traded for height, but I am not sure if I can find a good way of expressing the energy transfer from the wind to the aircraft. Another problem is that with non-constant airspeed, the drag will change and hence there will be energy transfer independent of the gradient.

IMHO, the airspeed is constant during the climb... But it's just an intuition...

The air speed is only constant if you manage to balance all forces with the change in wind speed. Without having any excess thrust that is only possible at one particular airspeed which depends on the wind gradient. And if you fly at that airspeed, the rate of climb can be anything you like. You just harvest the energy needed for climbing from the wind.

Imagine that you are flying s/l with no wind. Now, things change abruptly and the plane is in a headwind with a positive gradient. You do nothing... The plane will keep a constant airspeed while climbing through that gradient, with no need for extra gas, since the necessary energy will be provided by the gradient, which will 'suffer' accordingly. The way I have tried to demonstrate it is, always in my opinion, OK, but there is a step which has only intuitive (and dimensional) support, and is where I introduce the gradient ß as the factor of time, converting energy into power...

If there is a sudden step increase in headwind, there is a sudden step increase in airspeed. The aircraft will no be at constant airspeed anymore from that moment since the longitudinal forces are unbalanced due to change in drag.

Well, I say that of 'a sudden change' just to illustrate my point. In fact, a plane 'immersed' in a continuous gradient would 'feel' no 'sudden forces' at all...

Let us continue this discussion in chat.